The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is __________ mm².
( Given, y=2.0 × 10¹¹ Nm⁻²)

Solution

Physics sir · 4 steps

Stored energy is half of stress times strain times volume, so write it in terms of A and solve.

1. Energy in a wire. For a wire stretched by Δ L, the stored energy is 1/2 × force × extension. Use F = (YA Δ L)/L for the force and the area becomes the only unknown.

   U = 1/2 (YA (Δ L)²)/L

2. Put in numbers. With Δ L = 2 cm = 0.02 m, (Δ L)² = 4 × 10⁻⁴ m². So 80 = 1/2 × (2 × 10¹¹ × A × 4 × 10⁻⁴)/20 = 2 × 10⁶ A.
3. Solve for A. A = 80/(2 × 10⁶) = 4 × 10⁻⁵ m². Since 1 m² = 10⁶ mm², this is 40 mm².
4. So. The cross-sectional area of the wire is 40 mm².

The common slip: 20. Dropping the factor 1/2 and writing U = YA(Δ L)²/L gives 20, not 40.

The idea to keep. Elastic energy is 1/2YA(Δ L)²/L: the half comes from the force growing from zero. Work in SI units and convert to mm² only at the end.

Answer from the NTA final answer key. Solution by PrepWiser.